anonymous
  • anonymous
9x^2(2x+7)-12x(2x+7)=3x(2x+7)(3x-4) ...at the end where did the (3x-4) come from?
Mathematics
jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
As far as I can tell; you typed it in there. :)
anonymous
  • anonymous
It's on the website where I found this openstudy group. I never answered it. It says that's the answer and I'm a bit confused.
amistre64
  • amistre64
You have 2 numbers that are the same (2x+7) and (2x+7); so factor those out and your left with: (2x+7) (9x^2-12x) does this make sense? Now, we can factor (9x^2 -12x) even further to get: 3x (x-4) ; it should be: (3x) (2x+7) (x-4). It would be a typo :)

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amistre64
  • amistre64
ack!! my mistake 3x(3x-4) is correct... do you know why?
anonymous
  • anonymous
No, I don't know why. xD I have a math 11 test tomorrow and I'm a bit nervous. I can't get the hang of this stuff. :/ Mind explaining?
amistre64
  • amistre64
do you understand the first step....how to factor out (2x+7) and get left with (9x^2 -12x)? It is the opposite of distirbuting...
anonymous
  • anonymous
The question is 9x^2(2x+7)-12x(2x+7) and I know that theres a 3x in each term and a (2x+7) so you can factor those out right? So on the site it says it equals 3x(2x+7)(3x-4) and I understand how I got the 3x(2x+7) because both terms have it, but I'm so lost as to where the 3x-4 came up.
amistre64
  • amistre64
lets take it step by step and you will see where it comes from...fair enough?
anonymous
  • anonymous
Kay. :)
amistre64
  • amistre64
9x^2(2x+7)-12x(2x+7) has a common factor that is blaringly obvious. It is (2x+7) right? just let me know if this is correct....
anonymous
  • anonymous
Yes, but doesn't it have 3x as well?
amistre64
  • amistre64
3x doesnt matter yet; it will come, but not yet. lets clean it up some first ok?
anonymous
  • anonymous
kk, so yes, 2x+7
amistre64
  • amistre64
when we factor out (2x+7) are we left with: (2x+7) (9x^2 -12x) ? yes or no...
amistre64
  • amistre64
think of (2x+7) call it A if you have to..
anonymous
  • anonymous
yes, we're left with it.
anonymous
  • anonymous
sorry i'm talkign long to reply. keep pressing enter instead of post.
anonymous
  • anonymous
taking.*
amistre64
  • amistre64
its ok... Now we can factor (9x^2 -12x). Do you see how we could do that?
anonymous
  • anonymous
Oh wait! So, you can take out a 3 because of 9 and 12, and you can also take out an x, hence 3x, and you can get a 4 because 3X4 = 12.
amistre64
  • amistre64
you are brilliant :)
anonymous
  • anonymous
So 3x(3-4)
anonymous
  • anonymous
So once we get that we just put the 2x+7 back in?
anonymous
  • anonymous
And our 3x would go in front.
amistre64
  • amistre64
3x(3x - 4) = 9x^2 -12x right?
amistre64
  • amistre64
but yes, you can put the numbers back in whatever order you feel comfortable with.
anonymous
  • anonymous
yes! :D
anonymous
  • anonymous
So that's why there was a 4. I was confused about that but now I get it, because of the 3-4, 3x4= 12. Makes so much more sense now.
amistre64
  • amistre64
Just remember to take things step-by-step and it should all work out :)
anonymous
  • anonymous
I'm still bad at this stuff, though. I understand that question, but there's like 4 different types of facorting we're doing in class and I get confused when I'm doing problems because I'm not sure which one I'm supposed to use.
amistre64
  • amistre64
I hate the way they teach factoring in school. Because there is only one method that you actually use.. it is to "un" distribute or take apart; what was already put together.
anonymous
  • anonymous
& my math teacher picks favorites, if you're not up there in your math grades then he'll get mad. It's really annoying. I failed my past two math tests. I had 88 on my first test, but when he started teaching I dropped down to half that mark, to like 44. Tomorrow's test is kinda crucial.
amistre64
  • amistre64
what factoring methods do you need help with?
anonymous
  • anonymous
I'm strong in every other subject besides math. Anyways, care to help me with a couple more problems?
anonymous
  • anonymous
I'll list a problem I was having trouble with today. One second, I'll find it in my book.
anonymous
  • anonymous
x^2+8x+12 x^2+3x-18 and it's supposed to be over each other.
anonymous
  • anonymous
The top part = (x+2)(x+3) right?
amistre64
  • amistre64
glorified fractions is all that is. The equation gives you clues as to what to do. Do you see the "sign" of the last term?
anonymous
  • anonymous
What sign? I'm not sure what you mean by that, sorry.
amistre64
  • amistre64
(+12) OR (-18) the last term is gotten by multipying 2 numbers together. What combination of "signs" when multiplied give you a (+) answer? What combination of "signs" when multiplied give you a (-) answer? Can you tell me?
anonymous
  • anonymous
two positives = positive, two negatives = positive, and a - and a + = a -, right?
amistre64
  • amistre64
thats right. So the last term gives us a clue about our middle term. Becuase the middle term takes those same 2 numbers and adds them together. Lets focus on the top: +8 +12. we need 2 numbers that are either (-)(-) or (+)(+) to get the (+12) right? and they need to add together to get a (+) for the middle term (+8). Does this make sense?
anonymous
  • anonymous
Yeah that makes sense, we're trying to get a LCD?
anonymous
  • anonymous
2 and 6?
amistre64
  • amistre64
we will soon enough; but lets take it step by step. We can start to set up our answer like this: Good now... (x 2 ) (x 6 ) what "sign" do we need to use to fill this in? to get a (+12) and a (+8)
anonymous
  • anonymous
+
amistre64
  • amistre64
Yes, (x+2) (x+8) is the answer for the top; store it away somewhere where it wont get broken...
amistre64
  • amistre64
opps... make that (x+2)(x+6) and feel free to correct me :)
anonymous
  • anonymous
That's the answer for the top? Don't the 6 break down to a 3? Or we can't do that because we're only going half of 12?
amistre64
  • amistre64
there is nothing that x and 6 have in common so it is as far as it will go. to double check, multiply them back together.
anonymous
  • anonymous
I understand.
amistre64
  • amistre64
x^2 +3x -18 the last term tells us what signs we need to use. and the middle term tells us which of those signs gets the bigger number. Does that make sense?
amistre64
  • amistre64
what makes a (-) when multiplied together?
anonymous
  • anonymous
+ and -
anonymous
  • anonymous
I don't know what you mean by which sign gets the bigger number, though.
amistre64
  • amistre64
good; then its our only option: (x+ )(x- ) is what we are dealt. now find 2 numbers that multiply to 18 and "subtract" to get 3. Do you see why we need the (+)number to be bigger than the (-)number to get (+3)?
anonymous
  • anonymous
I'm a bit confused.
amistre64
  • amistre64
the numbers are gonna be 6 and 3... to equal 18 right? -6 +3 = -3 +6 -3 = +3 do you see it?
anonymous
  • anonymous
I thought we were trying to equal 18? Not - and +3?
anonymous
  • anonymous
-3*
amistre64
  • amistre64
ok...we can do that as well; what are the factors of 18 that subtract to get (+3) for a middle term? 1 and 18 2 and 9 3 and 6 thats it... right?
anonymous
  • anonymous
yep
amistre64
  • amistre64
one number has to be (-) and the other number has to (+) In order to get the (+3) for the middle term, the largest number has to be (+) such as: +6 -3 = +3 and +6 times -3 equals -18 does this make sense?
anonymous
  • anonymous
yes, it does!
amistre64
  • amistre64
then we can fill in our stuff now: (x +6) (x -3) is our answer. for the bottom. Can we cross anything out from top to bottom that are alike?
anonymous
  • anonymous
x+6
amistre64
  • amistre64
Good, and we are left with: (x+2)(x+6) (x+2) ---------- = ----- (x+6)(x-3) (x-3)
anonymous
  • anonymous
so then it's gonna be x+2 over x+3
anonymous
  • anonymous
haha yep!
anonymous
  • anonymous
wait, -3
anonymous
  • anonymous
thats what I had written down on paper, typo on here. lol.
amistre64
  • amistre64
i wish I had time for more, but my macroeconomics class is starting and running for the next 2 hours. You can do this, it is really stuff you already know :)
anonymous
  • anonymous
Thanks for your help, and perfect timing. I have to go right now too.

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